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3 years ago

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Question 8 of 16

1 answer:

Korolek [52]3 years ago

5 0

Answer:

Explanation:

PE = mgh where m is the mass, g is gravity, and h is the height from which the object can potentially fall. We are given enough info here to solve for h. Filling in:

42 = 1.8(9.8)h and

$h=\frac{42}{(1.8)(9.8)}$ and rounding to 2 sig fig's as we need:

h = 2.4 m, choice D

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The Apollo Lunar Module was used to make the transition from the spacecraft to the Moon's surface and back. Consider a similar m

ivolga24 [154]

Answer:

v=6.05 m/s

Explanation:

Given that,

Th initial velocity of the lander, u = 1.2 m/s

The lander is at a height of 1.8 m, d = 1.8 m

We need to find the velocity of the lander at impact. It is a concept based on the conservation of mechanical energy. So,

$\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=W\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=F\times d\\\\\dfrac{1}{2}mv^2-\dfrac{1}{2}mu^2=mgd\\\\\dfrac{1}{2}m(v^2-u^2)=mgd\\\\v^2-u^2=2gd$

v is the velocity of the lander at the impact

g is the acceleration due to gravity on the surface of Mars, which is 0.4 times that on the surface of the Earth, g = 0.4 × 9.8 = 3.92 m/s²

So,

$v=\sqrt{u^2+2gd} \\\\v=\sqrt{(1.2)^2+2\times 9.8\times 1.8} \\\\v=6.05\ m/s$

So, the velocity of the lander at the impact is 6.05 m/s.

6 0

3 years ago

Help me please!!!!!!!!!!

Ksenya-84 [330]

Answer:

B because 17100 is the half of the carbon

7 0

3 years ago

what is the voltage in mv across an 5.5 nm thick membrane if the electric field strength across it is 5.75 mv/m? you may assume

inn [45]

Answer:

V = 31.62 m V

Explanation:

Given,

Voltage across thick membrane = ?

Thickness of the membrane, d = 5.5 n m

Electric field = 5.75 M V/m

we know

$E = \dfrac{V}{d}$

V = E d

V = 5.75 x 10⁶ x 5.5 x 10⁻⁹

V = 31.62 x 10⁻³ V

V = 31.62 m V

Hence, The Voltage across the membrane is equal to 31.62 m V.

3 0

4 years ago

A soccer ball kicked with a force of 12.5 N accelerates at 6.2 m/s’ to the right. What is the mass of the ball? Answer in units

stepladder [879]

Answer:

m = 2.01[kg]

Explanation:

This problem can be solved using Newton's second law which tells us that the force applied on a body is equal to the product of mass by acceleration.

$F =m*a$

where:

F = force = 12.5 [N]

m = mass [kg]

a = acceleration = 6.2 [m/s²]

$12.5=m*6.2\\m = 2.01[kg]$

3 0

3 years ago

Suppose an asteroid orbiting the sun had an orbital period of 7. 5 years. What would its orbital radius be?.

creativ13 [48]

By using the orbital period equation we will find that the orbital radius is r = 4.29*10^11 m

<h3>What is the orbital period?</h3>

This would be the time that a given body does a complete revolution in its orbit.

It can be written as:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }$

Where:

π = 3.14
G is the gravitational constant = 6.67*10^(-11) m^3/(kg*s^2)
M is the mass of the sun = 1.989*10^30 kg
r is the radius, which we want to find.

Rewriting the equation for the radius we get:

$T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }\\\\r = \sqrt[3]{ \frac{T^2*G*M}{4*\pi ^2} }$

Where T = 7.5 years = 7.5*(3.154*10^7 s) = 2.3655*10^8 s

Replacing the values in the equation we get:

$r = \sqrt[3]{ \frac{(2.3655*10^8 s)^2*(6.67*10^{-11} m^3/(kg*s^2))*(1.989*10^{30} kg)}{4*3.14 ^2} } = 4.29*10^{11 }m$

So the orbital radius is 4.29*10^11 m

If you want to learn more about orbits, you can read:

brainly.com/question/11996385

7 0

2 years ago